From 84cf61be1fe78258b2ffabefa385363e208911af Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Radoica=20Dra=C5=A1ki=C4=87?= <r.draskic@student.tudelft.nl>
Date: Sun, 9 Feb 2020 19:52:16 +0000
Subject: [PATCH] Add solutions to Einstein and Debye model.
---
mkdocs.yml | 5 +--
src/2_debye_model_solutions.md | 76 ++++++++++++++++++++++++++++++++++
2 files changed, 78 insertions(+), 3 deletions(-)
create mode 100644 src/2_debye_model_solutions.md
diff --git a/mkdocs.yml b/mkdocs.yml
index f087c1b9..4e7f7ee3 100644
--- a/mkdocs.yml
+++ b/mkdocs.yml
@@ -32,9 +32,8 @@ nav:
- Fermi surface periodic table: 'fermi_surfaces.md'
- Extra exercises: 'extra_exercises.md'
- Solutions:
- - Lecture 0: '1_einstein_model_solutions.md'
- - Lecture 1: '1_einstein_model_sol.md'
- - Lecture 2: '2_debye_model_sol.md'
+ - Einstein model: '1_einstein_model_solutions.md'
+ - Debye model: '2_debye_model_solutions.md'
theme:
name: material
diff --git a/src/2_debye_model_solutions.md b/src/2_debye_model_solutions.md
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@@ -0,0 +1,76 @@
+# Solutions for lecture 2 exercises
+
+### Exercise 1: Debye model: concepts.
+
+1, 2, 3.
+
+Look at the lecture notes.
+
+4.
+
+$$
+g(\omega) = \frac{dN}{d\omega} = \frac{dN}{dk}\frac{dk}{d\omega} = \frac{1}{v}\frac{dN}{dk}.
+$$
+
+We assume that in $d$ dimensions there are $d$ polarizations.
+
+For 1D we have that $N = \frac{L}{2\pi}\int dk$, hence $g(\omega) = \frac{L}{2\pi v}$.
+
+For 2D we have that $N = 2\left(\frac{L}{2\pi}\right)^2\int d^2k = 2\left(\frac{L}{2\pi}\right)^2\int 2\pi kdk$, hence $g(\omega) = \frac{L^2\omega}{\pi v^2}$.
+
+For 3D we have that $N = 3\left(\frac{L}{2\pi}\right)^3\int d^3k = 3\left(\frac{L}{2\pi}\right)^3\int 4\pi kdk$, hence $g(\omega) = \frac{3L^3\omega^2}{2\pi^2v^3}$.
+
+### Exercise 2: Debye model in 2D.
+
+1.
+
+Look at the lecture notes.
+
+2.
+
+$$
+E = \int_{0}^{\omega_D}g(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right)d\omega = \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\beta\hbar\omega_D}\frac{x^2}{e^{x} - 1}dx + T \text{ independent constant}.
+$$
+
+3.
+
+High temperature implies $\beta \rightarrow 0$, hence $E = \frac{L^2}{\pi v^2\hbar^2\beta^3}\frac{(\beta\hbar\omega_D)^2}{2} + T \text{ independent constant}$, and then $C = \frac{k_BL^2\omega^2_D}{2\pi v^2} = 2Nk_B$. We've used the value for $\omega_D$ calculated from $2N = \int_{0}^{\omega_D}g(\omega)d\omega$.
+
+4.
+
+In the low temperature limit we have that $\beta \rightarrow \infty$, hence $E \approx \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\infty}\frac{x^2}{e^{x} - 1}dx + T \text{ independent constant} = \frac{2\zeta(3)L^2}{\pi v^2\hbar^2\beta^3} + T \text{ independent constant}$. Finally $C = \frac{6\zeta(3)k^3_BL^2}{\pi v^2\hbar^2}T^2$. We used the fact that $\int_{0}^{\infty}\frac{x^2}{e^{x} - 1}dx = 2\zeta(3)$ where $\zeta$ is the Riemann zeta function.
+
+
+### Exercise 3: Different phonon modes.
+
+1.
+
+$$
+g(\omega) = \sum_{\text{polarizations}}\frac{dN}{dk}\frac{dk}{d\omega} = \left(\frac{L}{2\pi}\right)^3\sum_{\text{polarizations}}4\pi k^2\frac{dk}{d\omega} = \frac{L^3}{2\pi^2}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)\omega^2
+$$
+
+$$
+E = \int_{0}^{\omega_D}g(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right)d\omega = \frac{L^3}{2\pi^2\hbar^3\beta^4}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)\int_{0}^{\beta\hbar\omega_D}\frac{x^3}{e^{x} - 1}dx + T \text{ independent constant}.
+$$
+
+2.
+
+Note that we can get $\omega_D$ from $3N = \int_{0}^{\omega_D}g(\omega)$ so everything cancels as usual and we are left with the Dulong-Petit law $C = 3Nk_B$.
+
+3.
+
+In the low temperature limit we have that $C \sim \frac{2\pi^2k_B^4L^3}{15\hbar^3}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)T^3$. We used that $\int_{0}^{\infty}\frac{x^3}{e^{x} - 1}dx = \frac{\pi^4}{15}$.
+
+### Exercise 4: Anisotropic sound velocities.
+
+$$
+E = 3\left(\frac{L}{2\pi}\right)^3\int d^3k\hbar\omega(\mathbf{k})\left(n_B(\beta\hbar\omega(\mathbf{k})) + \frac{1}{2}\right) = 3\left(\frac{L}{2\pi}\right)^3\frac{1}{v_xv_yv_z}\int d^3\kappa\frac{\hbar\kappa}{e^{\beta\hbar\kappa} - 1} + T \text{ independent part},
+$$
+
+where we used the substitutions $\kappa_x = k_xv_x,\kappa_y = k_yv_y, \kappa_z = k_zv_z$. Finally
+
+$$
+E = \frac{3\hbar L^3}{2\pi^2}\frac{1}{v_xv_yv_z}\int_0^{\kappa_D} d\kappa\frac{\kappa^3}{e^{\beta\hbar\kappa} - 1} + T \text{ independent part} = \frac{3L^3}{2\pi^2\hbar^3\beta^4}\frac{1}{v_xv_yv_z}\int_0^{\beta\hbar\kappa_D} dx\frac{x^3}{e^{x} - 1} + T \text{ independent part},
+$$
+
+hence $C = \frac{\partial E}{\partial T} = \frac{6k_B^4L^3T^3}{\pi^2\hbar^3}\frac{1}{v_xv_yv_z}\int_0^{\beta\hbar\kappa_D} dx\frac{x^3}{e^{x} - 1}$. We see that the result is similar to the one with the linear dispersion, the only difference is the factor $1/v_xv_yv_y$ instead of $1/v^3$.
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